Chemical Engineering - Fluid Mechanics - Discussion
Discussion Forum : Fluid Mechanics - Section 1 (Q.No. 31)
31.
The drag co-efficient for a bacterium moving in water at 1 mm/s, will be of the following order of magnitude (assume size of the bacterium to be 1 micron and kinematic viscosity of water to be 10-6m2/s).
Discussion:
24 comments Page 3 of 3.
Vicky said:
4 years ago
Re = (0.001*10^-6)÷10^-6 = 0.001.
Re<0.2.
So 24/Re = 24/0.001=24000.
Re<0.2.
So 24/Re = 24/0.001=24000.
(1)
Amar said:
4 years ago
24000 is the correct answer.
(2)
Athraa said:
3 years ago
Here, Re =1000.
TANMAY SHAH said:
2 years ago
Nre = d * v * ρ / μ
d = 1 micron = 10^-6 meter
v = 1 mm/s = 10^-3 m/s
ρ = 10^3 kg/m3
μ = 10^-6 × 1000 = 10^-3 kg/m.s
[ because; dynamic μ = kinematic μ × ρ ]
Nre = 10^-6 × 10^-3 × 10^3 /10^-3.
Nre = 10^-3 = 0.001.
Now for Nre < 1 it is called laminar flow region.
In the laminar flow region, called the Stokes law region the drag coefficient,
Cd= 24 µ/d ρ ν = 24 / Nre
Cd = 24/0.001 = 24000.
So, option-A is the right answer.
d = 1 micron = 10^-6 meter
v = 1 mm/s = 10^-3 m/s
ρ = 10^3 kg/m3
μ = 10^-6 × 1000 = 10^-3 kg/m.s
[ because; dynamic μ = kinematic μ × ρ ]
Nre = 10^-6 × 10^-3 × 10^3 /10^-3.
Nre = 10^-3 = 0.001.
Now for Nre < 1 it is called laminar flow region.
In the laminar flow region, called the Stokes law region the drag coefficient,
Cd= 24 µ/d ρ ν = 24 / Nre
Cd = 24/0.001 = 24000.
So, option-A is the right answer.
(10)
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